How many terms of the A.P. `-6`, `-(11)/2,\ -5 ...`are needed to give the sum `-25`?

Options

A. `5,20`
B. `5,10`
C. `6,12`
D. `6,10`

Answer

`A`

Explanation

Here, `a=-6, d=(-11)/(2)-(-6)=(1)/(2)`
Let -25 be the sum of n terms of this A.P. (n`in` N)
Using `S_(n)=(n)/(2)[2a+(n-1)d]`
`rArr -25=(n)/(2)[2(-6)+(n-1)((1)/(2))]`
`rArr -50=n(-12+(n-1)/(2)) rArr -50=n ((n-25)/(2))`
`rArr -100=n^(2)-25n rArr n^(2)-25n+100=0`
`rArr (n-5)(n-20)=0`
`:. n=5,20`
Both the values of n are natural and therefore , admissible.
Explanation of Double Answer :
`S_(20)=-6-(11)/(2)-5-(9)/(2)-4-(7)/(2)-3-(5)/(2)-2-(3)/(2)-1-(1)/(2)+0+(1)/(2)+1+(3)/(2)+2+(5)/(2)+3+(7)/(2)`
`=-6-(11)/(2)-5-(9)/(2)-4 " " `(other terms cancel)
` = S_(5)`